Finding the Coefficient of x^6y^3 using Binomial Theorem in (3x-2y)^9

[reenmachine]

Homework Statement

Use the binomial theorem to find the coefficient of ##x^6y^3## in ##(3x-2y)^9##.

Homework Equations

##1+9+36+84+126+126+84+36+9+1##

(I used two lines for the lenght)

##1(3x)^9(-2y)^0+9(3x)^8(-2y)^1+36(3x)^7(-2y)^2+84(3x)^6(-2y)^3+126(3x)^5(-2y)^4##
##+126(3x)^4(-2y)^5+84(3x)^3(-2y)^6+36(3x)^2(-2y)^7+9(3x)^1(-2y)^8+1(3x)^0(-2y)^9##

(again using two lines because of the lenght)

##19683x^9 - 118098x^8y + 314928x^7y^2 -489888x^6y^3 +489888x^5y^4 -326592x^4y^5##
##+145152x^3y^6 - 41472x^2y^7 +6912xy^8 -512y^9##

So the coefficient of ##x^6y^3## would be ##-489888##

any help will be greatly appreciated! thank you!

 

[mfb]

You can check your result with WolframAlpha.
I don't see where you would need help.

 

[reenmachine]

You can check your result with WolframAlpha.
I don't see where you would need help.

thank you! I wasn't aware of that site

 

[micromass]

There's really no need to write out the full polynomial. You only need one term.

 

[reenmachine]

There's really no need to write out the full polynomial. You only need one term.

You're right , I guess since it was the first time I used the theorem without a perfect ##(x+y)^n## I wanted to verify it.

thank you!

 

[HallsofIvy]

The problem, according to you, said "Use the binomial theorem" and you did NOT do that.

The binomial theorem says that [itex](a+ b)^n= \sum_{i=0}^\infty \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}[/itex]. Here a= 3x and b= -2y. You want "the coefficent of x⁶y³ with n= 9.

So this is the "i= 6" term: [itex]\begin{pmatrix}9 \\ 6\end{pmatrix}(3x)^6(-2y)^3= \begin{pmatrix}9 \\ 6\end{pmatrix}(729)x^6(-8)y^3[/itex] so the coefficient is [itex]\begin{pmatrix}9 \\ 6\end{pmatrix}(729)(-8)= \begin{pmatrix}9 \\ 6\end{pmatrix}= 5832\begin{pmatrix}9 \\ 6\end{pmatrix}[/itex].

Of course, [itex]\begin{pmatrix}9 \\ 6 \end{pmatrix}= \frac{9!}{6!3!}= \frac{9(8)(7)}{6}= 3(4)(7)= 84[/itex].

 

[Theorem.]

Yes, as HallsofIvy pointed out, the whole point here of using the Binomial Theorem is that you DO NOT actually have to do a full expansion. You just need to pick the appropriate terms out of the summation!